Problem: In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.

Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.

When Takahashi eats symbols `+`, `-`, `>` and `<`, the sequence A and the integer P will change as follows:

* When he eats `+`, the value of A_P increases by 1;
* When he eats `-`, the value of A_P decreases by 1;
* When he eats `>`, the value of P increases by 1;
* When he eats `<`, the value of P decreases by 1.



Takahashi has a string S of length N. Each character in S is one of the symbols `+`, `-`, `>` and `<`. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?

Constraints

* 1 \leq N \leq 250000
* |S| = N
* Each character in S is `+`, `-`, `>` or `<`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the answer.

Examples

Input

5
+>+<-


Output

3


Input

5
+>+-<


Output

5


Input

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<


Output

475

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

typedef long long lint;
typedef long double louble;

template<typename T1,typename T2> inline T1 max(T1 a,T2 b){return a<b?b:a;}
template<typename T1,typename T2> inline T1 min(T1 a,T2 b){return a<b?a:b;}

const char lf = '\n';

namespace ae86
{
	const int bufl = 1<<15;

	char buf[bufl],*s=buf,*t=buf;

	inline int fetch()
	{
		if(s==t){t=(s=buf)+fread(buf,1,bufl,stdin);if(s==t)return EOF;}
		return *s++;
	}

	inline int ty()
	{
		int a=0,b=1,c=fetch();
		while(!isdigit(c))b^=c=='-',c=fetch();
		while(isdigit(c))a=a*10+c-48,c=fetch();
		return b?a:-a;
	}

	template<typename T> inline int ts(T *s)
	{
		int a=0,c=fetch();
		while(c<=32 && c!=EOF)c=fetch();
		while(c>32 && c!=EOF)s[a++]=c,c=fetch();
		s[a]=0;
		return a;
	}
}
using ae86::ty;
using ae86::ts;

const int _ = 250007;
const lint bas = 250007 , mo = 2059392230891;

inline lint sfx(lint a,lint b)
{
	a%=mo,b%=mo;
	return (a*b-(lint)((louble)a/mo*b+0.5)*mo+mo)%mo;
}

inline lint powa(lint a,lint t)
{
	lint b=1;
	while(t){if(t&1)b=sfx(a,b);a=sfx(a,a),t>>=1;}
	return b;
}

inline lint inva(lint a)
{
	return powa(a,mo-2);
}

const lint ibas = inva(bas);

int n;char s[_];
lint loc[_];
unordered_map<lint,int> cnt;

inline lint trans(lint h,char op)
{
	if(op=='<')return sfx(h,ibas);
	if(op=='>')return sfx(h,bas);
	if(op=='+')return (h+1)%mo;
	if(op=='-')return (h-1+mo)%mo;
	throw "fuck me";
}

int main()
{
	ios::sync_with_stdio(0),cout.tie(nullptr);
	
	n=ty(),ts(s+1);

	loc[0]=1;
	for(int i=1;i<=n;i++)
	{
		loc[i]=loc[i-1];
		if(s[i]=='<')loc[i]=sfx(loc[i],ibas);
		if(s[i]=='>')loc[i]=sfx(loc[i],bas);
	}

	lint tar=0;
	for(int i=n;i>=1;i--)tar=trans(tar,s[i]);
	cnt[0]=1;
	lint ans=0,now=0;
	for(int i=n;i>=1;i--)
		now=trans(now,s[i]),ans+=cnt[sfx(now-tar+mo,loc[i-1])],cnt[sfx(now,loc[i-1])]++;
	cout<<ans<<lf;
	
	return 0;
}